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natima [27]
2 years ago
9

Find the next two terms of the following sequence: 14,38,74,122,182,254

Mathematics
1 answer:
Roman55 [17]2 years ago
8 0

ANSWER

The next two terms are

338, 434

EXPLANATION

The given sequence is 14,38,74,122,182,254

Let us observe some pattern and use it to find the next two terms.

14+24=38

38+36=74

74+48=122

122+60=182

182+72=254

To get the next term we add 84 to 254

254+84=338

To get the next term,we add 96 to 338

338+96=434

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Because you can only add and subtract radicals with the same radicand,
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Please help! I don't understand how to solve this problem
Ilia_Sergeevich [38]

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

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The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

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  • n is the sample size.
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Assuming an uniform distribution, the standard deviation is given by:

S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

M = z\frac{\sigma}{\sqrt{n}}

0.006 = 1.96\frac{1.1547}{\sqrt{n}}

0.006\sqrt{n} = 1.96 \times 1.1547

\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}

(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2

n =  142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

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